Question

The Fourier series of the function,
f(x) = 0, $-\pi <x\le 0$
$=\pi -x,0<x<\pi$
in the interval $[-\pi ,\pi ]$ is

f(x) $=\frac{\pi }{4}+\frac{2}{\pi }[\frac{\cos x}{1^2}+\frac{\cos 3x}{3^2}+....]+[\frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+....]$ The convergence of the above Fourier series at x = 0 gives

• A. ${\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}$
• B. ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n^2}=\frac{{\pi }^2}{12}$
• C. ${\sum }_{n=1}^{\infty }\frac{1}{(2n-1{)}^2}=\frac{{\pi }^2}{8}$
• D. ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{2n-1}=\frac{\pi }{4}$

Answer 1

The correct option is C

${\sum }_{n=1}^{\infty }\frac{1}{(2n-1{)}^2}=\frac{{\pi }^2}{8}$

At point of discontinuity, convergence of Fourier series is obtained by using the result $\frac{f\left({\alpha }^{-}\right)+f\left({\alpha }^{+}\right)}{2}$
So at x = 0:
$f\left(0^{-}\right)=li{m}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}\left(0\right)=0$

$f\left(0^{+}\right)=li{m}_{x\to 0^{+}}f\left(x\right)=li{m}_{x\to 0}(\pi -x)=\pi$

So at x = 0:
$f\left(0\right)=\frac{f\left(0^{-}\right)+f\left(0^{+}\right)}{2}=\frac{0+\pi }{2}=\frac{\pi }{2}$

So putting the value x = 0 in given Fourier sereies expansion,
$f\left(0\right)=\frac{\pi }{4}+\frac{2}{\pi }[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....]+0$
$\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }{\sum }_{n=1}^{\infty }\frac{1}{(2n-1{)}^2}$
or ${\sum }_{n=1}^{\infty }\frac{1}{(2n-1{)}^2}=\frac{{\pi }^2}{8}$